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virtus

find the equation of the locus of a point that moves so that its distance from the line 3x-4y+9=0 is always 5

  • one year ago
  • one year ago

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  1. ganeshie8
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    let that point be P (x, y)

    • one year ago
  2. ganeshie8
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    distance from a point to a line = \(\frac{Ax+By+C}{\sqrt{A^2 + B^2}}\)

    • one year ago
  3. ganeshie8
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    can you put the equation for distance ? from point P(x, y) to line 3x-4y+9=0

    • one year ago
  4. virtus
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    oh so it would be (3x-4y+9)/square root( 3^2 +(-4)^2)

    • one year ago
  5. virtus
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    (3x -4y+9)/5 =5 3x -4y+9 =25 3x-4y -16 =0

    • one year ago
  6. virtus
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    is that right ganeshi8?

    • one year ago
  7. virtus
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    btw i thought there was an ABSOLUTE SIGN around Ax + By + C for the formula

    • one year ago
  8. ganeshie8
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    yep! u r right.. we must get two lines

    • one year ago
  9. ganeshie8
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    |(3x -4y+9)|/5 =5

    • one year ago
  10. ganeshie8
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    3x-4y+9 = 25, and, 3x-4y + 9 = -25

    • one year ago
  11. virtus
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    ngawww i get it !

    • one year ago
  12. ganeshie8
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    it tricked me.... i was thinking there should be two lines at the very end. nice catch :)

    • one year ago
  13. virtus
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    thanks again - LOL and about writing GAPS yesterday, it wasn't meant to be rude or anything, just expressing my gratitude for your help ;)

    • one year ago
  14. ganeshie8
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    lol fine i get it nw :))

    • one year ago
  15. ganeshie8
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    |dw:1341791339629:dw|

    • one year ago
  16. ganeshie8
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    two lines we just got... which are at a distance of 5 from the given line... they look as above

    • one year ago
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